text:
Why does elevator trim hold a given airspeed?
I'm trying to understand a principle described in the book Stick and Rudder by Wolfgang Langewiesche between angle of attack and airspeed in relation to the elevator trim. He states that as angle of attack increases, speed decreases (given a constant power setting), and vice versa. In this section of the book, he's talking about this in relation to the use of the elevator trim for maintaining a specified airspeed. The book is focused on giving pilots practical explanations of how to fly, but is a little lacking and outdated on some scientific explanations as it was written in 1944.
My question is, what is a more scientific and aerodynamic explanation for why the change of AOA of the elevator (trim on elevator) determines the speed of the aircraft in a stabilized condition? What is it about the different camber of the airfoil that affects how fast the aircraft will fly?
I understand the practical use, I just want to understand why it works.

Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed.

Quite a few links to highly related questions/ answers were removed by the move. Go to the chat (https://chat.stackexchange.com/rooms/143195/discussion-on-question-by-rattmuscle-why-does-elevator-trim-hold-a-given-airspee ) to see those linked related questions/ answers.

The math and physics theory behind the phrase "pitch for speed, throttle for rate of climb" is what you're after.  First, let's talk about what it means.
In a video game or in your first intuition, it may make sense to 'nose up go up, nose down go down'.  Likewise, your intuition and simple video game may behave by 'more throttle is faster, less for slower'.  But that isn't how airplanes really fly in equilibrium (steady) states.
"Pitch for speed" means that controlling pitch (usually via trim) is the pilot's primary control for airspeed.
"Throttle for rate of climb" means that changing the throttle is the primary pilot's control for rate of climb.
In level equilibrium (steady, level flight, the state we spend most of our time in), Lift is equal to Weight, and Thrust is equal to Drag.
In a steady climb, we have (where $\theta$ is the climb angle and $RoC$ is the rate of climb).
$L=W \cos (\theta)$
$RoC = V \sin (\theta) = (T-D)V/W$
We can derive these equations if you'd like, but we can answer your question with just the steady-level flight version (i.e. $\theta$=0) $L=W$.  The rest is needed to explain 'throttle for rate of climb'.  Which is likely to be your next question.
Next, we write the lift equation in coefficient form.
$C_L = \frac{W \cos (\theta)}{q\,S}$
Where $q=0.5 \rho V^2$, the dynamic pressure.  And $S$ is the wing reference area.
Let me know if you don't know where the definition of lift coefficient comes from.  For now I'm assuming that you already understand that.
We solve that equation to put $V$ on the LHS -- just to emphasize it.
$V = \sqrt{ \frac{2\, W \cos (\theta)}{C_L\, \rho\, S}}$
Of course, starting with the level flight version...
$V = \sqrt{ \frac{2\, W}{C_L\, \rho\, S}}$
This equation was derived based on 1) Summing forces and setting them equal to zero (Newton's laws) and 2) The definition of lift coefficient (how aerodynamic forces change with density, size, and airspeed).  That is it.  This equation must be true (in steady-level flight) unless you find fault with Newton's laws or the definition of lift coefficient.
How does this answer our question?  How does this lead to "pitch for speed"?
You want to change (control) the aircraft's equilibrium airspeed -- $V$.  To do that, while keeping this equation true, you must change something on the right-hand-side to make it true.  What on the RHS does the pilot have control over?  Let us consider our options.
$2$ - The pilot is unable to change the value of two.
$W$ - The pilot has some control over the weight of the aircraft at takeoff.  And during flight, fuel burns reducing the weight slowly -- but not adequately to 'control' airspeed.  The pilot could throw payload overboard (reducing speed), but there is no practical way to gain weight during flight.  Weight is an unsatisfactory way for the pilot to control airspeed.
$S$ - Some high lift devices can be considered to change the wing area (though engineers would likely leave the reference area unchanged) -- and a morphing wing might be able to change wing area in flight.  So someday, wing area may be a control for airspeed.  However, for most conventional aircraft today, the pilot can not change the wing reference area in flight.
$\rho$ - The air density changes with altitude and with temperature (hot or cold day conditions), but a pilot that wants to fly at a given altitude (say 10,000 ft.) does not have the ability to change the density of air.
That leaves us with $C_L$, the lift coefficient.
$C_L$ - The lift coefficient is controlled by the pilot by either the elevator, or the elevator trim.  Since we are talking about equilibrium, let's focus on the trim change.  Changing the trim changes the lift coefficient (and angle of attack) where the pitching moment on the aircraft is zero.  (If you don't understand the details of trim, let us know, we can explain that too.)
By decreasing $C_L$ (nose down trim, decrease pitch), we increase speed.
By increasing $C_L$ (nose up trim, increase pitch), we decrease speed.
There you have it, "Pitch for Speed".
We can do the same thought experiment with the rate of climb equation, but it is already in the form we need...
$RoC = (T-D)V/W$
At a given moment in flight, if the pilot wants to change rate of climb, what can they do to the RHS of the equation to make it happen?
The only satisfactory control variable is Thrust -- which is controlled by throttle.
"Throttle for Rate of Climb".
One fun thing about this is to imagine that instead of a pilot operating a given aircraft, you are an aircraft designer.  The equations are the same, but your perspective on them changes...
$V = \sqrt{ \frac{2\, W}{C_L\, \rho\, S}}$
Now perhaps you want the aircraft to fly efficiently at a given flight condition ($V$, $\rho$).
Perhaps by 'efficiently', you mean that you want to fly at the best lift-to-drag ratio.  This occurs at a specific value of lift coefficient, we'll call it ${C_L}^*$.
That leaves the aircraft designer with control over the wing loading $W/S$ -- one of the most important aircraft design parameters.  It is the scaled version of "how big is the wing".
Similarly, the other main aircraft design parameter is "how big is the engine".  The scaled version of that is the thrust to weight ratio "T/W".  If you look back at the rate of climb equation and imagine yourself an aircraft designer, I think you'll see the importance of $T/W$.
Let us know if you need further explanation of any of the steps I skipped.

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My question is, what is a more scientific and aerodynamic explanation
for why the change of AOA of the elevator (trim on elevator)
determines the speed of the aircraft in a stabilized condition? What
is it about the different camber of the airfoil that affects how fast
the aircraft will fly?

This answer will tackle the problem from the view point of stick-fixed stability, i.e. assuming that the elevator trim wheel position completely determines the actual position of the elevator (or the all-moving horizontal tail).  That's admittedly a bit of an oversimplification.
For the aircraft to be in equilibrium, net pitch torque must be zero.  For a given angle-of-incidence of an all-moving horizontal tail, or a given elevator position of a "conventional" horizontal tail, the downlift or uplift created by the tail, and the lift created by the wing, can only create equal and opposite pitching moments around the C.G. when the wing is at one particular angle-of-attack.  At any other angle-of-attack, the pitch torques from the wing and tail won't be in balance, and the aircraft will pitch up or down.  If the aircraft has been designed to have good stick-fixed stability, this pitching effect will bring the wing to the angle-of-attack where the pitch torque from the wing exactly balances the pitch torque from the tail.1  So the position of the elevator trim determines the position of the elevator, and the position of the elevator determines the angle-of-attack of the wing.
By setting the wing's angle-of-attack, we are also setting the airspeed.  The reason for this is that for any given wing, lift is determined by angle-of-attack and by the square of airspeed, and total lift must equal weight, or else the flight path will curve up or down.  Any upward or downward curvature of the flight path will make the weight vector become partly parallel to the flight path, so that it acts like a thrust vector or a drag vector, driving a change in airspeed.2
Actually, the statement that "total lift must equal weight" is only true for wings-level, horizontal flight.  For a given angle-of-attack, banking actually causes the airspeed associated with any given angle-of-attack of the wing (and therefore with any given elevator position) to increase3, because lift must be greater than weight.  And in a wings-level climb or descent, there's actually a very slight decrease4 in the airspeed associated with any given angle-of-attack of the wing (and therefore with any given elevator position), because the total lift vector must equal weight times the cosine of the climb angle or descent angle.5
For more detail, see various answers to the related ASE question Does “trimmed for current speed” mean that the plane will do whatever to maintain the current airspeed? , especially this one.
Footnotes:

An excellent resource for better understanding the "balancing act" between the tail and the wing, and why a given elevator position will only "balance" the wing at one particular angle-of-attack of the wing, is John S. Denker's excellent "See How It Flies" website.

When the airspeed, and therefore the lift vector, are "too high" or "too low" for the angle-of-attack of the wing, the complete dynamics by which the airspeed changes and eventually settles down to the "correct" value for the angle-of-attack are beyond the scope of this answer.  Typically--in the absence of some "help" from the pilot or autopilot-- a series of slowly damping pitch "phugoid" oscillations are involved, with the flight path alternately curving upward and downward until the aircraft settles into a new flight path, with the climb or descent angle "correct" for the power setting, and the airspeed "correct" for the angle-of-attack.  To a first approximation, angle-of-attack can be viewed as remaining constant throughout the pitch phugoid, but the airspeed (and therefore the lift vector) is constantly "out synch" with the value that would be required to maintain linear flight at any given instant.  Also the direction of the weight vector relative to the direction of the flight path is constantly changing, so the thrust-like or drag-like effect from the weight vector is constantly changing.

To a first approximation, the increase in trim airspeed associated with banking is in proportion to the square root of the increase in wing loading--so for shallow bank angles, it will be minor.  Actually, the increase in trim airspeed is somewhat greater than given by this formula, because the curving nature of the relative wind in turning flight "bends" the relative wind upward at the tail, so that the tail's angle-of-attack relative to the local relative wind is more positive or less negative than it would be in linear flight, which shifts the pitch torque generated by the tail in the nose-down direction and causes the wing to trim to a slightly lower angle-of-attack.  (This can also be understood as a "pitch damping" phenomenon, i.e. an aerodynamic resistance to pitch rotation.)  This trim change associated with turning flight is covered to some extent in the resource linked in footnote 1.  That's why glider pilots often hold a lot of aft stick input when circling in a thermal updraft-- even if the glider is trimmed to fly at the min. sink angle-of-attack when wings-level.

Re "very slight" -- the effect is "very slight" for typical light general aviation airplanes, because typical climb or glide angles are so shallow, and may well be completely over-shadowed by other pitch trim effects relating to the thrust line.  In practice, reducing power to enter a descent may well cause a slight increase in airspeed.  However, in a very-high-performance airplane with a thrust-to-weight-ratio exceeding one-to-one, we certainly wouldn't expect the same relationship between control stick position and airspeed in horizontal flight at full power, as in a steady-state sustained vertical climb at full power!  See footnote 5 for more.

For more see Does lift equal weight in a climb? and other answers to related ASE question "Does lift equal weight in a climb?"  Also, it may seem extremely counterintuitive to imagine that angling the flight path down into a dive would ever decrease the airspeed--after all, the weight vector is gaining a (thrust-like) forward component, relative to the flight path.  However, keep in mind that in the context of this question, which assumes a given position of the elevator trim, i.e. a given position of the elevator, and therefore a given angle-of-attack of the wing, we are causing the flight path to descend by reducing power, or by lowering the landing gear, or by deploying some other sort of speed brake (or drogue chute?) that doesn't affect the wing's lift coefficient.  We're not talking about simply moving the control stick or yoke forward to reduce the angle-of-attack of the wing.

Thank you so much for this answer, I honestly found it (and all of the related articles you shared) to be the most useful in bridging the gap between my previous understanding to the more complex and mathematical explanation in the answer I chose. I had a hard time choosing between your answer and his but eventually settled on his as it explains it to an even greater detail. But I really appreciate the help you provided in giving me the steps to understanding the topic. Much appreciated and have a great day!

In my other answer, I assumed you understood how trim works.  Perhaps that is incorrect, so let's talk about pitch trim.
The airplane lift coefficient is modeled as:
$C_L=C_{L,0}+C_{L,\alpha}\,\alpha+C_{L,\delta}\,\delta$
Where $\delta$ is the control deflection.  And $C_{L,\alpha}$ is the derivative of the lift coefficient with respect to $\alpha$, and the other control derivatives are similar. $C_{L,0}$ is the lift coefficient at zero alpha and zero elevator deflection.
And the pitching moment coefficient is similarly modeled as:
$C_M=C_{M,0}+C_{M,\alpha}\,\alpha+C_{M,\delta}\,\delta$
The trim problem is solving the above two equations and two unknowns simultaneously.  We want to find $C_L$ equal to some desired target value -- ${C_L}^*$, and CM equal to zero.  We can do this manually, but we usually set up as a matrix problem...
$C_L={C_L}^*$
$C_M=0$
$\begin{bmatrix}C_{L,\alpha}&C_{L,\delta}\\C_{M,\alpha}&C_{M,\delta}\end{bmatrix}\begin{bmatrix}\alpha\\\delta\end{bmatrix}=\begin{bmatrix}{C_L}^*-C_{L,0}\\-C_{M,0}\end{bmatrix}$
This is a static trim question -- we're looking at the steady equilibrium value of angle of attack and control deflection required to achieve any desired lift coefficient at zero moment.  It ignores the oscillations and time dynamics that occur when you change from point to point.
When the pilot changes the control deflection (or the trim tab position), they move the airplane to a new trim solution -- i.e. a different angle of attack where moment is zero.  There is a new $C_L$ that corresponds to that new trim position.
In this set of equations, the only thing the pilot had direct control of is the elevator angle $\delta$.  So, by controlling $\delta$, they control $C_L$, and per my other answer, by controlling $C_L$, they control speed.

After reading all three of the answers and provided sources and links, I feel like I finally have a deep understanding of the greater picture of the relationship of AoA, trim, and airspeed. Thank you for the mathematical explanation and time spent on spelling it out for me. It is much appreciated! I also apologize for my delayed response, been a busy day. Thanks again and have a great day!

@RattMuscle -- the concept of ASE is that it's fine to take time craft a good answer, and similarly, a rapid response in terms of upvoting etc is not necessarily expected.  Comments are considered entirely peripheral and a "thank you" is not expected, much less a rapid one, so no need for apologies of any kind.  But I'm sure all appreciated your kind words all the same--


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Can I log dual received if I received training from a CFI and I sit in the back seat?

(h) Logging training time.

A person may log training time when that person receives
training from an authorized instructor in an aircraft, full flight
simulator, flight training device, or aviation training device.

The training time must be logged in a logbook and must:
(i) Be endorsed in a legible manner by the authorized instructor; and
(ii) Include a description of the training given, the length of the
training lesson, and the authorized instructor's signature,
certificate number, and certificate expiration date.

(14 CFR § 61.51)
Consider a 4-seat airplane with flight controls for the front seats only.  If there are two students on one flight and they go up with a CFI and neither one touches the flight controls can they both log the instruction time?  Assume the CFI takes off and shows the students how to conduct various maneuvers and how to track an ILS glide slope. Can the student in the back log training time since they received training from a CFI in an airplane? I sent off a request for legal interpretation but they said it could take up to four months to get a response.

What kind of training time do you mean? Do you mean flying hours?

If sitting in a passenger seat qualified as experience the world would be full of terrible pilots.

Would you add up the dual received column in your log book and count it as pilot time for the purpose of hitting the PPL hour minimums?

61.51 (c)Logging of pilot time. The pilot time described in this section may be used to:
(1) Apply for a certificate or rating issued under this part or a privilege authorized under this part. So PIC, SIC, instrument time, training time, NVG. Some call training time dual received but that is wrong it is "training time". 61.51 (h) states that training time is time obtained from instruction in an aircraft.

I would add up this training time for the purposes of obtaining a certificate.

If that worked, there would be an awful lot of people getting most of their dual instruction credit sitting in the back to save big bucks.  You're just an observer, not a student when you sit in the back.

It just say to log training time I have to be in an aircraft. The instructor has to be at the flight controls

@James -- you'd be lucky if you got a legal interpretation 4 months after your submission, I think the average is at least double that.  To get a sense of this, peruse the interpretations available here -- https://www.faa.gov/about/office_org/headquarters_offices/agc/practice_areas/regulations/interpretations/ .  Note also the content indicating that the FAA has tightened their policy and will now only issue interpretations on matters "that present a novel or legally significant issue" -- if they consider your request too trivial, they'll just ignore it --

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Can new thermals be generated above a cloud layer?
Thermals rise from the ground due to uneven heating from the sun, then stop rising once they start condensing into clouds. My question is can new thermals be formed at the top of this cloud layer? I often see multiple layers of clouds at the same time, are these higher layers formed by thermals or some other process? If there are thermals above the first cloud layer, are glider pilots able to make use of them?

Thermals get much stronger once they climb into the clouds!

Normally thermals used by glider pilots are driven by heating of the ground.

Thermals rise from the ground due to uneven heating from the sun, then
stop rising once they start condensing into cloud.

That's not quite the right "spin" on things.  The top of the cumulus cloud does mark the point where the thermal has "run out of steam" and can go no higher.  But the fact that the thermal has reached the point where its water vapor has started condensing into cloud does not have any sort of damping influence on the thermal.  In fact, the condensation process actually releases heat into the air that helps the thermal rise higher than it otherwise would have.  This is due to the "latent heat of condensation".  This plays an important role in the formation of thunderstorms and hurricanes.

I often see multiple layers of clouds at the same time, are these
higher layers formed by thermals or some other process?

Normally, by some other process.  These higher clouds do not normally indicate the presence of thermal updrafts that would be useable by glider pilots.
However, one can sometimes notice that high-level stratus clouds can be heated by the sun enough to cause little cumulus-like "turrets" to form on their upper surfaces.  While not indicating lift that would be useable to glider pilots, this is still an interesting phenomenon.
This link on "Castellanus" clouds may be of interest.
The sky is complicated!
PS I'm assuming that you are not envisioning that a glider pilot might actually be towed up into a cloud or high alongside of a cloud and then proceed to fly on instruments to gain additional altitude inside the cloud after release from tow.  (As far as I know, this is never done, but flying gliders in clouds is in fact still legal in some countries.)  If such a technique were envisioned, the scope of the question of which clouds indicate "useable" lift would be considerably broadened.  See in particular the link on Castellanus clouds above-- it appears that these clouds contain rising air, but the rising air often does not extend to below the level of the cloud.

As far as I know, this is never done. Very unlikely these days. But it was decently common many decades ago.

@MartinArgerami -- that sentence you quoted was aimed specifically at the idea of towing into cloud or towing above cloud base alongside a cloud, not cloud flying in gliders in general.  Perhaps yours was too, I can't quite tell.

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How does mass affect rate of turn?
After seeing effect of speed and bank angle on rate of turn, how would one mathematicly calculate the effect of changing mass.  For example, in a 30 degree turn at 100 knots, how would the rate of turn be affected if identical planes, one at 10,000 lbs, one at 9000 lbs were compared.  Granted, the pitch and power settings may be slightly different, generating different amounts of lift.  Would the heavier plane turn faster due to its greater horizontal lift component, or would it's heavier mass keep it more in the same direction?

The second law of motion states that: $F = m * a$
Rate of turn will be determined by the acceleration: $\frac{F}{m} = a$
If your vertical lift is 10,000 lb, in a 30 degree bank, your lateral force is $10,000 * tan(30)$, your acceleration is $\frac{10000 * tan(30)}{10000}$, or $tan(30)$.
If your vertical lift is 9,000 lb, in a 30 degree bank, your lateral force is $9,000 * tan(30)$, your acceleration is $\frac{9000 * tan(30)}{9000}$, or $tan(30)$.
Thus you can see the rate of turn depends on the bank angle, not the mass.

You have stumbled on an intriguing and fundamental aspect of Physics, the equivalence of Gravitational Mass and Inertial Mass.
Gravitational mass is the "m" in the equation that is used to determine the "force" of gravity due to one mass M, acting on another body with mass m.
$F=G\frac{m M}{r²}$
Inertial mass is simply the mass in the famous equation based on Newton's second Law, F=ma, that is used to calculate the acceleration that results from applying a force F to a body with mass m.
The fact that these two completely unrelated (or what seem to be unrelated) masses, are identical, and, (to the degree that physicists have been able to measure them), are always identical, was and had been a deep mystery for a long time. This mystery, (which even Newton was confounded by and unable to explain), was only explained by Einstein, when he published his General Theory of Relativity in 1915. The explanation requires a more elaborate explanation than can be provided here, but if you're interested, check out this link.
But this equivalence is the answer to your question. As the mass of an aircraft increases, the "force" of gravity, and the lift required to counteract that force increase by exactly the same amount as the force required to generate the same sideways acceleration (turn rate).

It's going to be identical (in the ideal case).
I'm presuming that you want to maintain a constant-altitude, constant-speed, constant-bank turn.  To do that, the vertical component of lift must be equal to weight, or 1g.  Therefore a fixed bank will give you a horizontal component of lift that is a fixed ratio of the vertical lift.  For 30 degrees, this is 0.57 times the vertical component or 0.57g horizontally.
Increased mass(weight) requires additional lift for flight.  And that additional lift balances the inertia of the increased mass.
Obviously for a plane with no other changes, this increase in required lift will cause it to reach limits at a lower bank angle as the mass increases.


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Yawing during turns
This is a question about aeronautics. Does an airplane yaw while it is turning?

If it doesn't yaw then the nose points always in the same direction. If it has turn, say, 180° then it has yawn 180°

Welcome to Aviation.SE! Can you please clarify what exactly you are asking about? Are you talking about adverse yaw?

Perhaps OP asks if rudder input is given...

"Yaw" is used in several different ways in aviation.  In the strictest sense of the word, your question would appear to be essentially "when an aircraft is turning, is the yaw rate other than zero?"  Is this what you mean?  And if so, how could it not be?

Rolling vs. yawing, and ailerons use vs. rudder use are explained in Are the functions of ailerons and rudder similar?, as well as what is a coordinated turn.

Does an airplane yaw while it is turning?

Not always.
A turn is a curvature in the flight path.
Imagine we initiate a bank to the left in a sailplane, without touching the rudder.
As the aircraft rolls left, the nose will swing (yaw) to the right due to adverse yaw, and the yaw string will deflect to the right.
As the left bank angle increases, in most (but not all) gliders at some point the lateral force from the banked wing will overcome the opposing sideforce from the airflow hitting the side of the fuselage, and the flight path will start curving to the left.
As the left bank continues to increase, at some point the aircraft's inherent "weathervane stability" will act to prevent the sideslip angle (i.e. the yaw string deflection angle) from becoming any larger.  If the sideslip angle (i.e. the yaw string deflection angle) is not increasing, and the flight path is curving to the left, the aircraft must also now be yawing to the left, not to the right.
At some instant in time before the sideslip angle stopped increasing, it must have been the case that the sideslip angle was still increasing, but at a rate that exactly matched the increasing rate of leftwards curvature in the flight path.  At that moment in time, the aircraft's heading must have been momentarily constant.  At that moment in time, the aircraft was yawing neither to the right nor to the left.
So at that moment in time, the aircraft was turning without yawing.
A simpler (but less often-seen, except at airshows) case of turning without yawing is a 90-degree-banked "knife edge" turn, with the pilot holding top rudder so that the airflow strikes the earthward side of the fuselage and supports the aircraft's weight.  Some yaw rotation was involved in establishing the "knife-edge" attitude, but once it is established, the aircraft can fly around in circles indefinitely without any rotation about the yaw axis.

+1:  Making the distinction between Yaw and Sideslip is important.  I would perhaps have made this point a bit more emphatically. Perhaps "Yaw" string is a misnomer. It should probably better be called a "Sideslip" string.


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What is this single seat aerobatic plane?
We saw this plane this week in NW Arkansas, curious what it is?

Welcome to Av.SE!

If you snapped any more photos of it, please add them to the question.  Especially from different angles or showing more detail.

Please describe its maneuvers.  Why have you called it aerobatic?

Another 20 feet lower, and I would call it kindling.

Btw looks like I made an action re a proposed edit that I was being prompted to review, essentially simultaneously as someone else accepted the proposed edit.  Not trying to play round-and-round games here intentionally-- ; )

Do you have a really good zoom lens, or was this thing flying far closer to you than it should've been?  Normally you don't see planes flying this close to each other unless they're military or doing some kind of air-show, i.e., they're flying together deliberately and communicating with each other.

@DarrelHoffman -- what makes you think two planes were involved?

@quietflyer We have a photograph of a plane, from above, seems self-explanatory.  Unless the photographer was on a tall mountain maybe?

@DarrelHoffman - I would think a point on an adjacent ridge, somewhat lower than elevation of ridge behind airplane (since far horizon cannot be seen beyond ridge behind airplane), but above actual elevation or altitude of target airplane, would be the perfect vantage point for this photo-- i.e. the airplane appears to be  flying down a valley slightly below the photographer's eye level, while banking toward the photographer --

Note that all the trees are equally in focus.  Suggesting a photo w/ a telephoto lens. No reason to assume that this plane was within inches or feet  of dragging its landing gear through any particular tree tops.

@quietflyer Yes, perspective and lens choice makes reality very hard to discern from that angle.  When I saw the photo, I just immediately imagined one of those infamous "quick jumping trees" taking that cute little bird down.

It's a GB-1 Game Bird built by Game Composites.

This picture shows the two place configuration but the OP's photo shows the single place option which is why the nose looks longer.
Here is a single-place version:

It is indeed an unlimited class aerobatic aircraft.

The canopy looks rather different - barely reaches the trailing edge of the wing in the OP's photo.

@MikeB I'm not sure where you're looking, but the canopy extends pretty far into the wing in the OP's photo. The single place GB-1 looks the same.

The specs at the Game BIrd page says it has a total fuel capacity of 81gallons.  Not knowing anything about small planes - how far does that get you? (Doesn't sound like much.) 'Cause that page also says the plane is "the best touring airplane in the world".

@davidbak Around 1000 nautical miles, according to the link at the top of the answer.

@davidbak And with a cruise speed of 200kn, that equates to 4 minutes per gallon, sounds reasonable.

@MikeB -- note the completely different perspective (viewing angle) in the two photos of the single-seaters.  The aircraft geometry looks the same to me.  Just as a (swept-wing) aircraft with dihedral will appear to have anhedral when viewed  from certain angles, and vice versa--


text:
How can ground speed be higher than airspeed?
I've heard that ground speed is higher than air speed when the plane is being pushed by wind wind. But how is this possible? Wouldn't your AirSpeed also increase since you're going faster?

If you walk towards the front of the train, your ground speed is greater than your
TrainSpeed, right? Same thing for airplane in the atmosphere, when the atmosphere is moving across the ground.

Have you ever watched a leaf floating along in a river?  Waterspeed = zero, yet it is moving relative to you standing on the shore.
"Wind" is just a mass of air, (or river of air) moving along the surface of the earth.
Consider the same comparison but with a hot air balloon; the airspeed of the balloon is zero, but it moves along the ground at the same speed as the wind.
Now consider powered flight - airspeed will be whatever it is, but your speed and track over the ground will depend on which direction you are pointing relative to the direction the airmass is moving along the surface.
Does that make sense?

It does help a bit.

Einstein was right: everything is relative.
Ground speed is speed relative to the ground, while airspeed is relative to the local air volume that the aircraft travels in. A spec of dust blowing in the wind has zero airspeed, but we see it flying past while we look at it with our feet on the ground.

Wouldn't your AirSpeed also increase since you're going faster?

Faster relative to the ground, yes. Not relative to the air.

Interesting, thanks!!

Planes fly through the air, not over the ground. The air itself is moving over the ground, which we call wind.  It may be easier to grasp this if you first think in terms of water, since you’ve probably seen (or even experienced) how that works.
Imagine you are in a boat on a river, and you are paddling at a fixed speed of 10 knots through the water. However, the water itself is moving at 5 knots relative to the shore.
If you paddle upstream, your boat’s ground speed is 10-5=5 knots, but if you paddle downstream, your boat’s ground speed is 10+5=15 knots.
Is the boat really moving three times as fast? That depends on whether you’re measuring its speed relative to the water or relative to the shore.

Hah!  all directly on point, except, to be picky, planes do fly over the ground... The point is that which frame of reference you choose to measure speed in is entirely arbitrary, and should be chosen based on your goal or the purpose you wish to use the "speed" for.  You measure it with respect to the ground to determine things that are associated with the ground, like enroute or arrival time, etc.  . You measure it with respect to the atmosphere for aerodynamic things, like stalls, or best climb rate, or gallons per hour fuel consumption, that are related to aerodynamics.

When determining how much force (or pressure) to inject into the aircraft carrier catapult, they have to calculate what the jet's speed will be relative to the boat, when it achieves the necessary (safe) airspeed for flight.

@CharlesBretana All true, but this is the simplest explanation I’ve found for people who don’t yet grasp wind as a moving air mass. Most default to a mental model of a car traveling at a fixed ground speed, with wind as a mysterious force pushing on the front or back of the car, and they won’t “get” the real answer until that mental model is replaced.

You have been here before I can see. Just a thought, for your consideration, but I think the hurdle for most people is the assumption that there is only one real frame of reference (the one they live in), or that the one they live in has and deserves some special privilege or preference over any others.  This is the unrecognized or unconscious assumption they have to be educated out of.  But planes DO fly over the ground...

@CharlesBretana I still struggle with that myself when it comes to relativity!

The really kool part of that is when you try to wrap the relativity concept around acceleration!  The Mach Principle (not yet proven as I understand it), is that there is no intrinsic fundamental frame of reference for acceleration either. The centrifugal force, and circular acceleration of an object rotating on a merry-go-round, in this view, is caused by the relative motion of all the other objects in the universe that are spinning around the object when measured in the merry-go-round's frame of reference.  Try to wrap your head around that!

The plane doesn't get "pushed" by the wind. It flies inside of the wind.
It's as if the plane's flight dynamics take place inside of a gigantic balloon.  You don't need to know whether the balloon is stationary over the ground in still air, or drifting over the ground at 150 mph in the jet stream.  The plane's flight dynamics in relation to the local airmass-- the air inside the balloon-- remain exactly the same.
Think about it for a moment. Don't you believe that groundspeed can be lower than airspeed when the plane is flying into the wind? So why shouldn't the reverse be true when flying with the wind?
It's like if you were running on one of those conveyer belts that you find in airports with long walkways.  Assuming for the moment that conveyer belt is also moving all the air above it along with it, so air resistance isn't a factor, if you want to cover 10 feet of belt per second, do your leg muscles have to work any harder if you are running one way on the belt, or the other way?  No.  So there you are, running along covering 10 feet of belt per second-- so let's say your "beltspeed" is 10 feet per second-- will your floorspeed or roomspeed-- the speed at which you are moving relative to the rest of the room-- be lower than 10 feet per second if you are running against the direction the belt is moving?  Sure.  And will your floorspeed or roomspeed be greater than 10 feet per second if you are running with the direction the belt is moving?  Sure.
Now change the problem a bit to say "I'm running with the muscle effort it normally takes me to cover 10 feet per second when running over stationary ground."  Your question is essentially "when I'm running with the direction the belt is moving, and am covering 20 feet through the room per second (because let's say the belt is moving at 10 feet per second relative to the rest of the room), does that boost in my "roomspeed" somehow also translate into a boost in my "beltspeed"?  And isn't the boost in my "beltspeed" somehow such that my "beltspeed" somehow magically becomes exactly equal to my "roomspeed"?
(This is analogous to the idea that "airspeed" (beltspeed) should not be lower than "groundspeed" (roomspeed), i.e. that "groundspeed" (roomspeed) should not be higher than "airspeed" (beltspeed).)
How could that make any sense?  And what number would you expect the "beltspeed" and the "roomspeed" to end up at, in a scenario like that?  It sounds like we are setting up a feedback loop where increased "roomspeed" drives increased "beltspeed" (so that it becomes equal to the "roomspeed" again) which drives still more increased "roomspeed" which drives still more increased "beltspeed"-- hey, I think we've just invented the warp drive!


text:
What Class Airspace Surrounds TRI Cities?
I'm planning out a "fly the Appalachian Trail" route, and as the AT comes out of the Great Smokey Mountains and heads for its crossing of the Nolichucky river, it puts anyone flying along it extremely close to KTRI which looks like this on a sectional:

What does the grey-outlined airspace from 3,500/4,000/5,100 to 10,000 ft around the airport signify?
If I look VERY closely I can see the Class D (D = Dashed) outline... but Class D airports don't have radar airspace...
It looks like Class C, but it's the wrong color (grey, instead of 'cranberry') and Class C don't have dashed outlines for their local control jurisdiction...
Since the route is meant to be flown VFR, it's important for me to know if I need clearance or just radio contact to traverse the airspace.

A sectional always has a Legend on the side of the paper map.  I took a screenshot of part of the Legend and attached it to the answer by 757toga.

@abelenky Thank you for that. I looked at it, but my eyes simply refused to find that.

@randomhead It answers the second half of it, yes. I'm fine with this getting marked duplicate if you want, although from a purely pedantic perspective it's not actually a duplicate question.

The first half can be answered by looking at the legend of a sectional.... you are also incorrect about "class D airports not having radar airspace." This is by no means always true.

The airspace at Tri-Cities Airport is Class D when the tower is open and surface-level Class E when the tower is closed.  The shape of this airspace is a simple circle, 4.3 miles in radius.  There is also a small surface-level Class E "extension" (depicted by a dashed magenta line) extending to the northeast, which is in effect 24/7.  The descriptions can be found on pages D-112, E-110, and E-206 of the Airspace Designations and Reporting Points document (FAA Order 7400.11F).
The grey line, and the elevation figures, denote the boundaries of a "Terminal Radar Service Area".  This is noted on the Sectional Chart Legend

(source: i0.wp.com)
and also on page 19 of the Aeronautical Chart User's Guide.
This information (as well as the actual schedule of the operating hours of the tower/ effective hours of the Class D airspace) is also available in the current (24Mar2002) Chart Supplement which was accessed via the FAA Digital Products Search Page.
The resource mentioned in another answer gives some good information on the practical significance of a TRSA to VFR pilots, and on why some TRSAs still exist at all rather than being converted to ARSAs (now Class C airspace) as was the FAA's intention at one time.
Here's a bit of closing trivia-- when TRSAs were first introduced, they were charted with solid magenta lines.  I've seen examples of this on sectional charts from 6-9-83 and 11-19-87.  ARSAs ("Area Radar Service Areas", now Class C airspace areas) were introduced later, and were originally charted with wide segmented blue lines (e.g. sectional charts from 11-21-85 and 11-20-86).  Then ARSAs were changed to wide segmented magenta lines (e.g. sectional charts from 11-19-87 and 8-20-92).  Then TRSAs were changed to solid grey-black lines (e.g. charts from 4-30-92 to present).  And then ARSAs (now Class C airspace) were changed to solid magenta lines (e.g. charts from 11-12-92 to present).  So TRSAs have been charted in two different ways, and ARSAs (now Class C airspace areas) have been charted in three different ways, and the current charting method for Class C airspace areas (formerly ARSAs) was once used for TRSAs!


text:
What advantage might interplane-strut mounting offer to ailerons, on the Curtiss Model F?
Glenn Curtiss's 1912-1918 Model F biplane flying boat has large ailerons on the struts between the wings, not contiguous with any other flying surface.  Did this aileron position have any advantage that either he knew of or that we now know?
[Per SupBruh's comment, contrast this position to conventional upper-wing position.  Lower wing was rare even then, and particularly for a seaplane where an aileron this big would hit the water when deflected.]
He knew about wing warping, so he knew the effectiveness of camber (hinging an aileron at a wing's trailing edge) as compared to just changing an airfoil's angle of attack.  He knew how to make a hinge: this aircraft uses them at rudder and elevator.
He also knew about modern ailerons: in 1908 he flew an aircraft with them.
Maybe he chose such a large chord for the ailerons to try to increase their authority.  Photos suggest that the airfoil was just a flat plate, which stalls at a deflection angle much less than that of a conventional aileron.  But the drag from a stalled aileron that big would produce scary adverse yaw.

At times, patent battles with the Wright brothers were an issue in where Curtis chose to place his ailerons.  I don't have time to type up an authoritative answer right now so maybe someone else can run with it-- but I see that you are not primarily asking about motivation.  But that being the case, you might consider changing the title to match what you really most want to know?

Disclaimer: I can only speculate, having no direct involvement with the development of the Curtiss Model F.
Stick forces were not really an issue with the small and slow early aeroplanes. As both size and speed went up, they would become a design consideration first because the pilot strength is lowest for lateral motion of the stick or control horn. With more aerodynamic experience, elaborated ways of reducing stick forces were employed.
My explanation, therefore, is that this is an early attempt at reducing stick forces. Hinged at the trailing edge, the complete chord length of the aileron would be behind the hinge line and no offsetting moment could be produced. Also, being part of the wing, ailerons hinged at the trailing edge will float up once lift is created. This requires both ailerons to be interlinked with stiff control cables which again raises stick forces, now from friction. Also, this preload from lift limits the effectiveness of a downward deflection of a wing-mounted aileron.
The free-flying aileron, on the other hand, has the advantage of starting with a clean sheet, if you will:

No preload exists, so it will be equally effective in up- and downward deflection.
It is not encumbered by the boundary layer of a wing flying before it, so its neutral axis will be easy to find and be constant over angle of attack, unlike that of a wing-mounted aileron. This was certainly not formulated this way at the time of the design of the Model F, but experience would bring the designers to exactly the same conclusion. The same considerations brought Junkers (more precisely: Otto Mader) to the offset flaps and ailerons in their designs from the late 1920s to early '30s.
With the aileron position between wings, the hinge line can be freely chosen at any chord position. The picture below (source) shows that a small part of the aileron was ahead of the strut, suggesting that the hinge line was not at its leading edge, but set back a bit. This enables the forward part to act as a balance, thus reducing stick forces.

I would expect that the Curtiss Aerorplane Company tried out different hinge line positions, which is easily done with the inter-wing arrangement, and settled on the one shown above, because it reduces stick forces considerably but still produces a clear centering force, which is important for agreeable handling qualities.
I do not think that the inter-wing position would reduce wing flex: Lift forces from the aileron are introduced into the wing by the rear strut, so both torsion and bending are very similar to a wing with ailerons on its trailing edge. To reduce torsion, the aileron would have to be mounted on the forward struts. Then they would even warp the wing in a way supporting the intended rolling moment. I speculate, however, that Curtiss and his engineers maybe considered that option but were afraid of the resulting aerodynamic interference between wings and ailerons.
That the inter-wing position lost against the balanced, wing-mounted aileron is certainly due to the much lower drag of the latter. Note that the inter-wing aileron needs its own rigging. With the increased experience of balancing ailerons, the wing-mounted version lost its main disadvantage and became the overall better solution.

"Disclaimer: I am not over 130 years old." :)

Interplane-strut mountings would offer multiple benefits:

Aerodynamic stability (it would reduce the chances of structural failure)
Structural stability (to put it simply, it would make the plane stronger)
reduce wing flex (it would greatly reduce the flexing of the wing caused by strong forces of, well, air)
render ailerons more useful (Interplane-strut mountings would keep the wing straight, allowing the ailerons to do their job properly. without interplane strut mountings, the wing would flex due to the upward force that the ailerons produce when you roll. When the wings flex, (specifically to top wings of this biplane) it will reduce airflow to the ailerons, rendering them much less useful. with interplane strut mountings, the wing will no longer be able to flex as much, allowing maximum airflow to the ailerons.

By stability I refer to both more aerodynamic stability and less wing flexing. interplane strut-mounting would also increase the planes critical velocity (enabling you to go faster without completely destroying your sweet ride) and with the wing warping, my mistake. with all the wires between the wings I had though for a moment that the airplane's primary form of control was wing-warping. Now that I remember what plane this is and after zooming in much closer, I can see that this plane is fully aileron-controlled and not partially aileron-controlled.

and my claim of the gain in structural/aerodynamic stability interplane strut mounting would provide, is because interplane strut mounting will form a triangle between the strut, the wing, and the fuselage. a triangle is one of the most structurally resilient shapes, due to the dispersion of tension on the shape (think suspension bridges.) It would reduce wing flexing from the force of the air against the aileron.

Thanks, that helps.  Because this is a Q&A site, not a chat room, could you move these explanations into your answer (under it, click Edit)?  That makes it easier for other readers to see the end result.  Also, if critical velocity means $V_{NE}$, the latter term is more recognizable.

sure thing (I edited the answer.)

and it also depends on specifically which wing(s) you're seeking to apply (or seeking to find the results of what improvements it would make if you applied) the mountings onto.

I disagree with your assertion that wing flex is reduced. After all, the aileron lift force is introduced into the wing by means of the strut. Bending and torsion will be the same as for a wing-mounted aileron. Besides, this wrong reasoning is used to support all four bullet points, so this answer is completely wrong.

It sounds like this answer is trying to address a different question, "what are the advantages of installing interplane struts on biplanes, including any advantages in terms of how well the (conventionally-positioned) ailerons will work".  Did you not realize that the actual question was more along the lines of "given that there are interplane struts, why put the ailerons there instead of on the trailing edges of the wings?" -?

Re " interplane strut mounting will form a triangle between the strut, the wing, and the fuselage" -- I fail to see how the specific items named here form a triangle. (But read here if you want to learn how other parts of the rigging of a biplane typically do form a triangle -- http://aviation.stackexchange.com/a/66468/34686 ).  Maybe what you are trying to say is that having an interplane strut allows the diagonal wire to complete a triangle whose other sides are defined by the strut and the wing, and whose apex is at the fuselage>

Both the answer and the comments by the answerer make more sense if we realize that he is treating "interplane strut mounting" as a synonym for "interplane strut".  (Either that, or he's using "interplane strut mounting" to refer to the diagonal bracing wires that run from the fuselage to the interplane struts.)  This answer has nothing to do with where the ailerons are mounted at all. He's just saying that the plane is more controllable if it is more rigid. And naturally, interplane struts (especially when combined with diagonal bracing wires) increase rigidity.

In fact the first comment states that when he originally conceived the answer, he didn't realize the plane had ailerons at all. And it's still not clear that he has yet realized that the ailerons are actually mounted on the interplane struts, and that that's what the question is actually about. Again, maybe a good answer-- but not to the present question. Hope something in these comments helps the answer to be edited to be improved.


text:
Entry to back-side of power curve
When entering slow flight, when do we enter the back side of the power curve? Is it when we reduce the throttle or when the airspeed decreases to just above 5 knots from the stall speed?

In a piston engine propeller powered aircraft, you're on the back side of the power curve when you are flying at lower than the speed for minimum power.

By definition, the power required to maintain steady level flight increases from this point whether we increase or decrease airspeed.  The important part here is that the power curve has a positive slope to the right (increasing airspeed) and a negative slope to the left (decreasing airspeed).
At speeds greater than minimum power required, the positive slope of power vs. velocity means that it takes more power to fly faster (this is normal).
At speeds slower than minimum power required, the negative slope of power vs. velocity means that it takes less power to fly faster (this is the back side of the power curve).


text:
Why Critical Angle of attack doesn't change with an increase of IAS/TAS?
I am trying to understand this topic that is not letting me sleep. I am a private pilot and I am studying for the CPL, if someone could give me a simple explanation of this I would be very grateful. Complex explanations are also welcome, thank you very much in advance.
I understand that a stall is an abrupt loss of lift due to the detachment of the boundary layer of an airfoil. As the angle of attack of an aircraft increases, it becomes more "curved" for the air that faces it, causing the air to no longer follow the shape of the wing profile and detach. The capacity of a fluid to follow the curvature of an object depends to a certain extent on its energy, and this can be seen with the use of Slots in airplanes, which re-energize the air flow and allow it to follow more curvatures. pronounced, and for this reason the critical angle of attack increases with the use of these devices. However, when we talk about fluid energy we are referring to its kinetic energy, which in theory could be increased by increasing the speed of the aircraft, so this would make the critical angle of attack to increase, but this does not happen. It is known that the critical angle of attack is an inherent property of the wing design and does not change with the speed of the aircraft. The real question then is why does the critical angle of attack not increase with increasing aircraft speed?

The standard explanation about how slots work is that they "re-energise" the airflow, whatever this means. But they simply work giving the following airfoil making up the flap a brand new boundary layer... as simple as that. This new boundary layer is more robust than the one coming from the wing.

But the slot/slat is a convergent duct, so the air exiting is slightly accelerated relative to the free stream.

Just to verify: you are asking why isn't the critical AoA (stall AoA) 20° at 200kts if it is 15° at 150kts (numbers just as an example). To me that is even more counterintuitive, since at a higher speed air molecules have to follow wing curvature faster, so they will experience higher acceleration. My original assumption when I started studies in this subject was that critical AoA should be smaller at higher speeds.

Basically it seems like another restatement of your question could be, if we put a wing test section in a wind tunnel and vary the wind speed, why should the airflow streamlines remain the same (within reason--i.e. we aren't changing the wind speed enough to cause huge changes in Reynolds number).  Especially near the stall a-o-a.

Related: https://aviation.stackexchange.com/q/51128/34686

@AlexGallegos not worth losing sleep over.  The understanding of the exact mechanism of slots is a very advanced subject.  Studies with jib sails over mainsail even suggest the slot mechanism slows airflow over the mainsail.  I believe the slot redirects the flow enough to delay separation.  With vortex generators, the effect is higher AoA before stall, but more drag.  For a GA aircraft it is way more important to know your stall and manuvering speeds, and the effect weight has on them.  Generally, in aviation, higher airspeed means lower AoA for the same lift.  Much safer.

We can see from an analysis of a Clark Y airfoil that lift is linear to Angle of Attack at angles of attack below stall.
Intuitively, this would mean that the higher suction, or lower pressure above the wing at higher airspeeds, would counter-act the increased momentum of the air molecules at higher airspeeds, keeping flow attached.
Beyond the scope of GA aircraft, higher Mach number will affect stall angles of attack, but for a plane that flies within the range of 50-150 knots, the old saw that stall AoA is always the same holds.
Note on the graph of coefficient of lift vs AoA, when approaching stall, lift is no longer linear to increasing AoA, indicating a collapse of the low pressure pocket above the wing leading to flow separation.
Also remember, for your exam, the bottom of the wing is still producing lift at stall AoA.

Well if you increase the airspeed you increase the lift and so will climb. This changes the direction of the relative airflow (it now comes more from above), thus slightly reducing your angle of attack.
Now you may say that a slot or slat will also increase lift and result in a climb, and this is true. But the pilot action is then to reduce the airspeed, which allows for an increase in angle of attack beyond the clean critical AOA.

You've complicated the picture by touching on pitch stability dynamics. This opens a whole other can of worms that we might explore further (phugoid etc) but might better simply be left unopened. Consider that if we've increased the airspeed and want to experimentally explore what the critical attack is now, one way to vary the angle-of-attack at will without climbing or pitching up into the start of a loop would be to slowly increase the bank angle while operating the controls (including throttle and elevator, but not flaps) as needed to hold both airspeed and altitude constant.

Re above -- the other condition I should have mentioned is that we should keep the aircraft in coordinated flight, so that the wing is doing all the lifting, not the side of the fuselage.  Also, at least for modest thrust/power levels, leaving the power or thrust setting fixed and allowing the a/c to climb or descend during the test would give essentially the same basic result, as long as the airspeed was held constant as the bank angle was increased.


text:
How to slow down while maintaining altitude
Is reducing throttle/power already enough? Or do I have to adjust the pitch as well? If so, I suppose pitching up as well? If pitching up is needed as well, is it because as speed decrease, there is less aerodynamic force acting on the tail and so the tail down force is less which pitch the nose down? Or is it because as speed decrease, lift decrease, so pitch up to create more lift?

To maintain altitude, lift must be equal to weight. The first question is, does slowing down reduce the amount of lift we have?
We can look at the lift formula for the answer:
$$\mbox{Lift} = C_l \times 0.5 \times \rho \times V^2 \times S$$
Airspeed is a part of this formula, represented by V. Therefore we can say that if we slow down, we will have a reduction in lift and therefore descend, if we don't change anything else.
So what can we change to maintain altitude? Working from the end of the formula: S can be increased with flap in some aircraft, but this is inefficient and cumbersome. Then, ρ is air density which is outside of our control. That leaves Cl, which is essentially a function of camber and angle of attack.
Again, camber could be increased with flap, but the best way to increase Cl is to increase our angle of attack. We can increase this via a pitch adjustment. So there we have our answer - if we want to maintain altitude while slowing down, we must either increase flap or, the preferred option, increase our angle of attack via an increased pitch attitude.

Consider the four forces acting on an aircraft in equilibrium:

Drag must exceed thrust in order to decelerate.  Presuming that you don't have speed brakes to increase drag, the only way to slow down while keeping lift and weight equal is to reduce thrust.
As the aircraft slows, the wing will produce less lift, requiring an increase in AOA to maintain lift equal to weight.  This will further increase induced drag, necessitating a power correction once the target airspeed is reached.

Re "necessitating a power addition once the target airspeed is reached."  -- not if initial and final speeds are on the "front side of the power curve".

@quietflyer, I changed to "power correction".  But... unless you know the exact power setting for your new target airspeed you will likely need to add more power than what was used to decel.  That's what I meant, not that the power would be higher than the first, faster speed.

If you reduce forward thrust then you implicitly reduce the lift generated by the wing in its current configuration. So to maintain altitude you must pitch up or otherwise change the configuration of the wing - which in itself will change the attitude.

Is reducing throttle/power already enough? Or do I have to adjust the
pitch as well?

It depends on the aircraft.  It also depends on whether you are flying on the "front side" or the "back side" of the power curve.
If you are flying on the back side of the power curve, you are going to need to increase power to maintain altitude after you slow to your desired airspeed.
As for pitch inputs--
Most aircraft, at least to a first approximation, have reasonably strong speed stability, meaning that they tend to maintain the trimmed airspeed, even as the engine power output is changed.  This is a secondary result of their tendency to maintain the trimmed angle-of-attack.  So normally, if we are trimmed for some given airspeed, we expect to have to exert some forward pressure on the stick or yoke to maintain a faster airspeed, and we expect to have to exert some aft pressure on the stick or yoke to maintain a slower airspeed-- especially if power is held constant.
But a closer look will reveal that many aircraft have some tendency to move toward a higher angle-of-attack (and a lower airspeed) as power is increased, and a lower angle-of-attack (and a higher airspeed) as power is decreased.  Most high-winged aircraft with tractor (front-mounted) engines fall into this category.
In such case, if you operating deep on the "back side" of the power curve, if you start out trimmed for level flight, it's conceivable that you might have to apply some forward pressure on the stick or yoke as you increase power after slowing down, to maintain your new target airspeed.
Some other aircraft--especially some with engines mounted on a pylon above the fuselage-- have some tendency to move toward a lower angle-of-attack (and a higher airspeed) as power is increased, and a higher angle-of-attack (and a lower airspeed) as power is decreased.
In such case, if you are operating on the "front side" of the power curve, if you start out trimmed for level flight, you'll likely have to apply some forward pressure on the stick or yoke in order to maintain your new target airspeed, after you decrease power to slow down.
But if you are just asking about the aircraft's pitch attitude-- slower flight is almost always associated with an increase in angle-of-attack.1  If the flight path remains horizontal, the aircraft's pitch attitude will always be shifted in the nose-up direction.
Footnotes:

Why "almost"?  Consider this "outlier" case-- "hovering" flight.  Google "model airplane 3-D hover" if you've never seen something shaped like a "conventional" airplane, with no rotor blades or vectored thrust capability, hang on the prop in a stationary zero-airspeed hover.  Here the pitch attitude is 90 degrees, but the angle-of-attack of the wing in relation to the total airflow (including the propwash) is zero degrees (or if the airfoil is non-symmetrical, whatever slight negative angle-of-attack is associated with zero lift.)  Nearly the same applies to flight at extremely low airspeeds, well below stall speed, where the plane is "creeping" slowly along in an extreme nose-high pitch attitude.  It appears to be the case that no full-sized aircraft lacking vectored thrust, rotors, etc can be flown in this manner at present.

"If you are flying on the back side of the power curve, you are going to need to increase power to maintain altitude as you slow down."  I'd suggesting rewording slightly to avoid reading like you need to add power to slow down. You must first decrease power to initiate the slow down, then add more power to maintain airspeed given the higher AOA needed to maintain altitude.

@MichaelHall -- thanks, done

Let me know what you think of my edit.  (again, the 4 forces... power counters drag, not weight)

@MichaelHall -- I decided to roll it back.  The edited statement is not necessarily incorrect, but I find it more natural to think of using pitch inputs to control airspeed and power changes to control vertical speed.  Especially on the "back side" of the power curve.

Ok, I agree when flying an approach, but all trimmed up during cruise do you really adjust throttle to hold altitude?  How much time do you really spend on the backside?

is reducing power enough, or do I have to adjust the pitch as well?

The pilot is generally ready to do both at the same time, but it helps to know pitch (angle of attack) and power requirements beforehand.
Dig into your Pilots Operating Handbook and find V min power.  This is the airspeed your plane uses the least amount of throttle in level flight (for a given flaps setting).
Any airspeed faster than this will require more power.
If your V min power for flaps zero (clean) is 60 knots, and airspeed is being reduced from 100 knots to 80 knots, you will simultaneously reduce power and pitch up to hold altitude, then re-trim for the new airspeed.

This is actually similar to what is done at the end of the downwind leg for recreational aircraft, where airspeed is reduced from pattern speed to approach speed.

Pitch for speed.  Throttle for rate of climb.
So, first you adjust your pitch trim to achieve the desired speed.  Then observe the rate of climb indicator.  If positive (most likely), then decrease throttle to reduce to zero.

Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed.

This is correct if one is flying on the backside of the power curve.  Here it is safer to pitch up first to desired airspeed.  Chopping throttle first in slow flight and pitching up can easily lead to a stall.


text:
FAA's Airplane Flying Handbook says more lift is necessary during a climb. Isn't it the opposite?
"If a climb is started from cruise flight, the airspeed gradually decreases is the airplane enters a stabilized climb attitude. The thrust required to maintain straight-and-leve flight at a given airspeed is not sufficient to maintain the same airspeed in a climb. Increase drag in a climb stems from increased lift demands made upon the wing to increase altitude. Climbing requires an excess of lift over that necessary to maintain level flight Increased lift will generate more induced drag. That increase in induced drag is why more power is needed and why a sustained climb requires an excess of thrust.
For practical purposes gravity or weight is a constant. A vector diagram shows why more lift is necessary during a climb, as the vertical component of lift generated from the wings is no longer perpendicular to the wings and adds to drag. The total vertical force is increased by adding a vertical component of thrust from the powerplant and the power should be advanced to the recommended climb power"
The above is what the FAA Airplane Flying Handbook says. However, I don't understand the bold statement. Why is more lift needed in a climb? Does it mean that more lift is needed to initiate a climb? Isn't it true that less lift is needed during a climb? Also I can't figure out how a vector diagram shows more lift is needed. I understand that the vertical component of lift is no longer perpendicular to the wings but it doesn't add to drag, does it? It is the horizontal component of weight that adds to drag, right?

I'm no expert on this sort of stuff, but isnt it a duplicate of https://aviation.stackexchange.com/questions/100378/is-more-lift-needed-for-a-stabilized-climb and/or https://aviation.stackexchange.com/questions/40921/does-lift-equal-weight-in-a-climb (if so please VtC as duplicate rater than answer again)

Also, to @hitomhi you've now asked basically the same question 3 times. At least 2 (I suspect soon to be 3) are marked as a duplicate, If you do think you have a question not covered already please distinguish why and stop just repeating the same question.

Depends on how you define "Lift". If you only include aerodynamic Lift from the wing, then the fact that the lift vector points slightly aft of pure vertical is of course relevant. But remember, that in level flight, the total aerodynamic force is also tilted backwards slightly!   If, otoh, you define lift to include the component of all forces (including thrust) that is normal to the earth, (purely vertical), then the answer to your question is, by definition, that the "Lift" in a steady state climb must be the same as the aircraft weight. It's all a matter of definition.

I've read https://aviation.stackexchange.com/questions/40921/does-lift-equal-weight-in-a-climb. It says less lift (only the force generated by the wing and is perpendicular to the wing) is needed during a climb. I understand but it opposes what the FAA says, right? since the FAA says more lift is needed. Or what does the "lift" mean in the FAA book? Correct me if I've made anything wrong. Thank you.

@Hitomhi: I agree that the text you cited is very unclear... From level flight you need a short "spyke" in lift to change the attitude of the airplane and start the climb. Once you've reached the needed climb angle you reduce again the lift but to a value which is lower than the one at level flight. Drag also reduces consequently. Thrust must be anyway increased because now it has to compensate for the relevant weight component (think about an airplane flying straight up). Let me know if this answers your question, I'll be glad to elaborate an answer if needed.

Yes, the FAA is wrong on this.  No big surprise; it's not the first time and it won't be the last.  See https://aviation.stackexchange.com/questions/40921/does-lift-equal-weight-in-a-climb/56476#56476 .  Thanks for bringing this particular FAA error to my attention.  I'm keeping a list and this is around number 20.

In addition to your own duplicates, the topic has been well covered here already.  I would encourage you to search other similar questions for a complete understanding.

@Hitomhi Please do not re-ask this question again. This material has been covered at length elsewhere, and further attempts to ask about the same concept aren't going to be received any better than your last couple of questions have been.  Thank you.

Added the tag "FAA-knowledge-test", as it seems to be the closest thing we have to FAA manuals or FAA pilot training materials.  Perhaps we need a new tag along these lines.

PS page number citations (along with a note as to which edition of the handbook you are referencing) would be helpful.

@quietflyer FAA-H-8083-3C page 3-19 paragraph 2 and 3.

Thanks, might consider adding that to answer, or when I get a chance later this evening I'll do it.

Appreciate that

"Increased drag in a climb stems from increased lift demands made upon
the wing to increase altitude."

False.

"Climbing requires an excess of lift over that necessary to maintain
level flight."

False.

"A vector diagram shows why more lift is necessary during a climb, as
the vertical component of lift generated from the wings is no longer
perpendicular to the wings and adds to drag."

False.

"The above is what the FAA Airplane Flying Handbook says. However, I
don't understand the bold statement. Why is more lift needed in a
climb? Does it mean that more lift is needed to initiate a climb?"

It is certainly true that the initiation of a climb is generally accompanied by a slight,temporary increase in lift force-- any bend (curve) in the flight path is a form of acceleration, and some force is needed to accomplish that-- but it's pretty clear that the FAA source is not trying to convey that particular nuance. The committee that wrote this book is clearly incapable of understanding such details. Rather, they are repeating the falsehood that more lift is needed in steady-state climbing flight, than in steady-state horizontal flight at the same airspeed.

Isn't it true that less lift is needed during a climb?

Yes.

Also I can't figure out how a vector diagram shows more lift is
needed. I understand that the vertical component of lift is no longer
perpendicular to the wings but it doesn't add to drag, does it?

Correct, it does not.

It is the horizontal component of weight that adds to drag, right?

Not exactly.  The component of weight that acts parallel to the flight path must be offset by thrust for steady-state flight.  In other words, making the simplifying assumption that thrust acts exactly parallel to the flight path, in steady-state climbing flight thrust must equal drag plus the component of the weight vector that acts parallel to the flight path.  So you could say that in climbing flight, the component of the weight vector that acts parallel to the flight path acts similar to drag, and must be offset by thrust, but doesn't actually increase the drag force.
For more, see various answers to the related ASE question Does lift equal weight in a climb?.
Take caution in treating FAA sources as authoritative in the matter of flight dynamics.  Sources written for basic pilot training are written by committee and tend to be riddled with errors.  See for example the confusing and incomplete diagrams addressed by this related ASE question: What is missing from these diagrams of the forces in slips and skids?.

@Hitohmi -- this answer might be expanded to note that we usually choose to fly at a increased lift coefficient when we want to climb, as opposed to cruising flight.  Which is another way of saying that we usually climb slower than cruise speed.  For any given speed, increasing the lift coefficient increases lift, and drag.  But we aren't free to hold airspeed constant and vary the lift coefficient.  Otherwise we'd end up flying loops, sooner or later.  Anyway you cut it, the FAA is wrong on this.


text:
Why the heavier the aircraft is, the higher the airspeed must be to obtain the same glide ratio?
I know weight does not affect glide ratio. But why does higher weight need higher gliding speed?

There are two ways to generate more lift, either fly faster or at higher AoA. Usually a high AoA would reduce your L/D.

In still air, the glide ratio of an aircraft is the same as its L/D ratio (more info here). The L/D ratio of a given aircraft depends only on its angle of attack; a given AoA will always correspond to one particular L/D ratio¹. So if a given AoA could be maintained, the L/D ratio (and thus the glide ratio) would be fixed.
Now, if the weight is increased, the lift required also increases. since we cannot increase the AoA (that would alter the L/D ratio), we must increase the airspeed to obtain the necessary lift.
1 Ignoring the effects of Mach number & Reynolds number

If AOA is unchangeable, how can I increase the airspeed in a glide? I will have lost my engine and thrust in that situation.

@Hitomhi -- this is the right answer.  I'm converting my explanatory comments, into an answer.

@Hitomhi for a given glide angle increased weight will increase airspeed.  Imagine a car on a hill, one is very light and one is heavily loaded.  The heavier one will coast down the hill faster.

@RobertDiGiovanni:  if we ignore the things we usually ignore-air drag, energy going int spinning the wheels, drag because the tire gets flattened where it hits the road, etc.  that is not true.  Both cars will go down the hill at the same speed.  The effect on the glider comes because we need more lift to counter the weight.

@RossMillikan ah, the feather and the lead ball in a vacuum.  This is aerodynamics.  An object with the same drag area will fall faster if it is heavier!  It's like this:  heavier glider --> falls faster --> produces more lift --> which supports its greater weight!  This will hold up to significant Mach effects.

@Hitomhi: maybe there is a misconception with the AOA and the plaines pitch. The AOA is the angle the air approaches the wing (and body) in relation to the planes length axis. The pitch is the angle between the planes length axis and the horizon. If the heavier plane keeps the same pitch it would have a greater AOA to created enough lift to support the planes weight. The heavier plane can establish the same AOA as the lighter plane by pitching down resulting in a higher speed.

The glide ratio depends only on the angle of attack during the gliding descent. Imagine two aircraft in a gliding descent: aircraft A is at its best glide speed, and aircraft B is at the same airspeed and angle of attack as aircraft A, but is heavier.
Since both aircraft have the same airspeed and angle of attack, they have the same lift and drag. In the lighter aircraft, the lift and drag are in equilibrium with the weight, and so the aircraft is in unaccelerated flight- it can keep the same angle of attack all the way to the ground without its airspeed changing.
Since the weight of aircraft B is larger, the forces are not in equilibrium and aircraft B's descent rate. This means a steeper glide path and the change in relative wind means a higher angle of attack if the pitch attitude is not lowered. Aircraft B could raise its angle of attack to arrest the descent rate and keep the same airspeed as aircraft A. But that would negatively affect its glide ratio. If it instead holds its angle of attack the same, it will continue to accelerate until its airspeed is sufficient to arrest the acceleration.

Both aircrafts have the same lift and drag? B is heavier, isn't the lift of B also higher so as to counteract the higher weight?

@Hitomhi Lift is a function of only airspeed and angle of attack. Indeed, for unaccelerated flight B needs a bigger lift than A. The only way it can get that lift without raising the angle of attack is to increase its airspeed.

If AOA is unchangeable, how can I increase the airspeed in a glide? I will have lost my engine and thrust in that situation.

Re "This means a steeper glide path and the change in relative wind means a higher angle of attack. " -- you are following a mental model where the pitch attitude is forced to remain constant.  It is a lot more intuitive to hold angle-of-attack constant.  (Because to a first approximation this is what happens when we hold control stick position constant-- or when we just let the glider fly at trim.)  Can you generate an explanation of why the heavier aircraft ends up speeding up to a higher airspeed, if angle-of-attack is held constant?  It's pretty straightforward...

@Hitomhi You don't have to do anything increase it. On the contrary, you would have to do something to keep it from increasing. The lift and drag at the lower airspeed are not sufficient to keep you in unaccelerated flight. If you just hold the angle of attack constant your plane can't help but accelerate until the increased drag and lift are enough to stop the acceleration.

@Chris Does it mean that speed will increase itself and ultimately remain at that airspeed?

@Hitomhi Yes. If you keep the angle of attack unchanged the airspeed will eventually reach an equilibrium point.

@Chris Got it thanks

In relation to this question, one might ask
"If AOA is unchangeable, how can I increase the airspeed in a glide?"
Let's assume that we are keeping the control stick fixed in the position that corresponds to the best glide angle in steady-state flight.  To a close approximation, this means that the Angle-of-Attack (AoA) is fixed.  Which-- ignoring Reynolds number and Mach effects-- means that whenever we are in steady-state flight with a fixed control stick position, the glide angle (and glide ratio) is fixed.
If the AoA is fixed, and the airspeed is too low for the present weight and AoA,, then both Lift and Drag will be too low. Meaning specifically that the sum of their components that act parallel to the flight path will be smaller than the component of Weight that acts parallel to the flight path. So the forces acting parallel to the flight path do not add up to zero, so there is a net force acting parallel to the flight path. This force will cause the airspeed to increase.
Meanwhile the forces acting perpendicular to the flight path also do not add up to zero, so the flight path curves earthward, which throws the forces acting parallel to the flight path further out of balance, exacerbating the increase in airspeed.
Eventually everything comes back into balance and the glider is back to the original glide angle at a higher airspeed. The entire process (which I've only described the first part of here) is called a pitch "phugoid". To a first approximation AoA is constant in a phugoid. That's not exactly the case in real life, but if it were, we'd still see an increase in weight lead to an increased glide speed, at the original glide angle. Study the pitch "phugoid" to learn more.
"But why does higher weight need higher gliding speed?"
See above-- that's why a higher weight leads to a higher gliding speed.  If you somehow increase the weight of the glider (perhaps by condensing moisture from the atmosphere to add to the on-board water ballast?), while holding the angle-of-attack constant, the airspeed must increase.

There's another phenomenon very similar to this. When you fix AoA but increase thrust, the airplane will accelerate and have excessive lift, hence fly up, then decelerated by  gravity. The end result is a climbing flight, but not faster.

@quietflyer Do you also mean the speed will increase itself like what Chris says above?

@Hitomhi Think of a falling feather. Both the glider and the feather have one commonality: both obtain their thrust from their own weight. If you were to increase the weight of the feather, it will have more "thrust", causing it to fall faster. The same is true for the glider.

I know weight does not affect glide ratio

That's correct.
The glide ratio is defined as the ratio (obviously) between the horizontal length flown by the aircraft divided by the relative lost height. Called $\gamma$ the relevant angle between these two components, the glide ratio simply equals $1/\gamma$:

(Picture source)
To prove this, we simply see from this picture that:
$D=W sin\gamma$
$L=W cos\gamma$
(Note that in the picture the weight is called $F_{gravity}$ while I've used the "standard" letter $W$)
If we now write the ratio $L/D$ we get:
$\frac{L}{D}=\frac{W cos\gamma}{W sin\gamma}=\frac{cos\gamma}{sin\gamma} \approx \frac{1}{\gamma}$
i.e. the ratio $L/D$ just equals the glide ratio! Furthermore, the weight $W$ has been cancelled out and therefore the glide ratio does not depend on the weight but only on the aerodynamic characteristics of the aircraft.

But if the weight is gone, how comes that

higher weight need higher gliding speed?

That's because when you glide you try to maximize the glide ratio i.e. you try to maximize $L/D$. And $L/D$ is maximised when you fly at one specific speed which does depend on the weight:
$V_{L/D_{max}}=K\sqrt{W}$
where $K$ is a constant calculated from $\rho$ and the aerodynamic polar of the aircraft and that, in your question, is the same for both the light and the heavy aircraft.

So yes, the glide ratio does not depend on the weight but the speed at which it is maximised does.


text:
How does the lift force stop increasing itself after a certain point?
I am currently trying to create a basic flight simulator and am very confused by the concept of lift. This is the equation of lift shown below that I found on the internet.
$$
F_l = \frac{1}{2} \rho v^2 C_l A
$$
It seems like the lift force would very quickly increase with airspeed (velocity). What I am unsure about is, what forces counteract lift to prevent it from reaching an extremely high value.
For other forces such as thrust in a plane, there tends to be a maximum force applied, so an opposing force such as drag cannot exceed it. But for lift, there doesn't seem to be one.
As the force of lift increases, won't it increase the velocity as a result, and through this increase in velocity, increase the force of lift and so on, until an extremely high value is reached? Of course, this will never happen in real life.
So what factors am I forgetting to account for that prevent this from happening?

How would an increase in lift increase velocity?

"As the force of lift increases, won't it increase the velocity as a result" -- why?  If I'm cruising level at 100 knots and I pull back the stick to start a loop, does my velocity start increasing?  What if I have a bunch of extra power in reserve and I open the throttle to exactly offset all the extra drag from the increased angle-of-attack -- in that case does the airspeed start increasing as the flight path curves upward, even though the weight vector is gaining a component that acts opposite to the direction of the flight path?  This question doesn't make sense.

If you roll 90 degrees and pull the stick all Lift goes into the turn.  But the drag profile of the aircraft in the direction of excess lift is much, much greater than head-on.  The longitudinal stability of the aircraft will try to point it in the direction of the new relative wind, ie, the turn.

Note that by definition, Fl is always perpendicular to v. If the airplane does gain some upwards velocity as result of lift, the airflow direction also changes.

It seems like the lift force would very quickly increase with airspeed (velocity). What I am unsure about is, what forces counteract lift to prevent it from reaching an extremely high value.

Let's look at your equation -
$$ F_l = \frac{1}{2} \rho v^2 C_l A $$
and lets rearrange to isolate the unknown variable -
$$ \frac{2F_l}{\rho v^2 A} = C_l  $$
Now let's discuss.  In level flight, $F_l$ is a constant equal to the known weight of the aircraft. And we know $\rho$ is a constant. The variable $A$, which is dependent on the aircraft, is also a constant.  But look... $v^2$ is there on the bottom and when $v$ gets larger, $C_l$ has to get smaller.  And why is that?  When the aircraft is in level flight, the force of lift $F_l$ - in other words, the weight - doesn't change. Consequently, only the velocity, and the lift coefficient, which is dependent on velocity, change. This is only true, tho, in level flight, for something called the static performance condition of the aircraft.  Increase the velocity, and the coefficient of lift will decrease in proportion to the square of the velocity. But the lift will remain the same.
Ok... So what changes the lift coefficient? As flight velocity increases, to keep lift the same, the angle of attack of the wing must be decreased.  Since the coefficient of lift is proportional to the angle of attack, a decrease in the angle of attack will proportionally decrease the coefficient of lift.
So the lift coefficient is also dependent on the angle of attack of the wing! That is why maneuvers with the aircraft are limited to velocities and force loadings within the performance envelope defined by the v-n diagram for that aircraft.  If the angle of attack is abruptly increased, the lift coefficient is abruptly increased, and the consequent increase in lift can exceed the load-bearing limit of the wing.  Nevertheless, more gradual increases in the lift coefficient can occur in level flight when the aircraft is banked.  Even tho the flight path may remain level and $v$ may remain constant, the increase in wing loading is inversely proportional to the cosine of the angle of bank, meaning that in level flight, as the bank angle increases, the wing loading increases. Why is that? The effective weight of the aircraft has increased in direct proportion to the inverse of the cosine of the angle of bank. Consequently, the angle of attack of the wing must be increased in response. What increased the effective weight of the aircraft? Well, centrifugal force, of course. The flight path of the aircraft, even tho level, is a circular arc. The centrifugal force is the horizontal component of the lift force produced in the bank. The vertical component is still equal to the weight of the aircraft. This answer may provide other insights.
So... everything is cool.  It's not seemingly complicated. We just need to understand what changes in the equation, and why.

What I am unsure about is, what forces counteract lift to prevent it from reaching an extremely high value

The airplane disintegrating is what would stop aerodynamic forces to grow too much.
That's why any aircraft has well defined speed limitations which must not be overcome. For example one of these speed limitations is called, quite meaningfully, "never exceed velocity".

This is the equation of lift:
$F_l = \frac{1}{2} \rho v^2 C_l A$
It seems like the lift force would very quickly increase with airspeed (velocity).

In that equation (which is correct) there are four terms, two of which can become zero more or less independently i.e. $v$ and $C_l$. So if $C_l$ is zero then the speed can become as big as possible but the aerodynamic force would remain anyway null.

I am currently trying to create a basic flight simulator and am very confused by the concept of lift.

Sorry but that sounds like "I'm building a vehicle and I'm very confused by the concept of wheel"
I'd suggest you to read some basic books about airplanes like Raymer's "Simplified Aircraft Design for Homebuilders". This NASA beginners guide to aerodynamics might also be helpful.

It seems like the lift force would very quickly increase with airspeed
(velocity).

That's right, provided that angle-of-attack stays constant.

As the force of lift increases, won't it increase the velocity as a
result

Why?
You have some confusion about basic physics.  The lift vector acts orthogonal (perpendicular) to the flight path and thus has no tendency to drive an increase in airspeed, generally speaking.
There are situations where the lift vector has an indirect tendency to drive an increase in airspeed.  One such situation is during the third quarter of a loop, starting from the point where we are exactly inverted.  The lift vector acts earthward and so at that instant, both the lift vector and the gravity vector are acting perpendicular to the flight path, curving (bending) the flight path earthward (downward).  As the flight path curves earthward, the gravity vector gains a stronger and stronger component acting parallel to the direction of the flight path, driving a rapid increase in airspeed, even in a glider where thrust is zero.

It seems like the lift force would very quickly increase with airspeed (velocity). What I am unsure about is, what forces counteract lift to prevent it from reaching an extremely high value.

It's the other way around: lift is used to counteract the force of gravity. To increase lift, you increase speed, angle of attack (AoA) or both. It is true that lift increases dramatically with speed, read further for more about that.

For other forces such as thrust in a plane, there tends to be a maximum force applied, so an opposing force such as drag cannot exceed it. But for lift, there doesn't seem to be one.

Drag certainly can and will exceed thrust. That is how airplames slow down. As stated above, the opposing force to lift is gravity. There is a limiting factor to lift, it is the critical AoA of the wing above which lift diminishes dramatically, and structural strength of the wings.

As the force of lift increases, won't it increase the velocity as a result, and through this increase in velocity, increase the force of lift and so on, until an extremely high value is reached? Of course, this will never happen in real life.

Again, it's the other way around. To increase lift, you increase speed, AoA or both. Speed is controlled by throttle, limiting factor being drag, and AoA is controlled by elevator, limiting factors being critical AoA and structural loads.
P.S. let's not drag how gravity is not a force into this

There are actually 2 opposing forces to lift: gravity (directly) and drag (indirectly).
Drag limits the speed one can go, thereby limiting the v$^2$ factor.
But if lift does exceed weight, flight path turns upwards.  Since the engine/prop is less efficient at producing vertical force than the wing, again speed is limited.  The thrust force must increasingly oppose the drag and weight, decreasing velocity.
One can see, in a vertical climb, the wing is useless.
Structural strength of the plane, G load limit, also limits amount of lift applied.  So, one can go into a high speed dive and apply maximum lift, only to have the wing spar fail.
And, of course, there are human pilot G load limits.
In the end, it is thrust performance and low drag that is desired, with just enough wing lift to carry the load.

Nothing forces it from increasing quickly. At high airspeeds an airplane cruises with a small angle of attack to reduce its lift. If it suddenly increases the angle of attack its lift will go up dramatically. This can cause the airplane to break apart mid-air.
You have some misconceptions about acceleration and velocity. Lift is always perpendicular to the airplane's flight path and so doesn't increase or decrease its speed- it causes the flight path to curve upward or downward.

Possibly the most useful single concept to understand is that of energy; the engine adds energy and it’s lost in parasitic and induced drag, anything remaining may ge converted to kinetic energy or potential energy.  To address your question directly, lift always has a component that’s in the opposite direction to thrust, so will always tend to slow the aircraft rather than speed it up.

"lift always has a component that’s in the opposite direction to thrust" -- why?

@Frog If you're taking about induced drag, that is not a component of lift. Lift is defined as the component of the net aerodynamic force which acts perpendicular to the flight path vector. Therefore, by definition, lift does not have a component opposite to thrust (assuming thrust acts in the direction of flight).

@quiet flyer - you’re right, it’s not invariably the case, but the OP seems worried that lift can somehow increase overall speed, which suggests that lift has a component in the same direction as thrust, although I doubt that they have considered the problem in quite those terms.

At its crudest, there are four forces in play that your simulator needs to deal with. Thrust, drag, lift, and gravity. And they all act on different parts of the aircraft. An imbalance arising from changing any of those forces results in a change in the velocity of the aircraft to reestablish balance, so yes, lift increases with forward velocity, but forward velocity is a consequence of balancing thrust and drag and since drag is not linear, high velocities are only possible with massive thrust, and with massive thrust, you can rocket to the moon.